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CaHuomwore |
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Answer» The efficiency of Carnot engine if T₁ and T₂ temperature are initial and final temperature . η=1-T₂/T₁ condition 1 :- when η ( efficiency of engine ) = 1/6 then, 1/6 = 1 - T₂/T₁ ---------(1) condition 2 : when temperature of sink is reduced by 65K then, η(efficiency of engine ) = 1/3 e.g, 1/3 = 1 - (T₂ - 65)/T₁ -----------(2) solve equations (1) and (2) 2(1 - T₂/T₁) = 1 - (T₂ - 65)/T₁2 - 1 = 2T₂/T₁ - (T₂ - 65)/T₁ = (T₂ + 65)/T₁T₁ = T₂ + 65 Put it in equation (1)1/6 = 1 - T₂/(T₂ + 65)1/6 = (65)/(T₂ + 65)T₂ + 65 = 390 ⇒ T₂ = 325K or 52°C now T₁ = 390K or 117°C |
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